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3 years ago

Use the drop-down menus to complete the statements to match the information shown by the graph.

Mathematics

2 answers:

posledela3 years ago

5 0

Answer:

Thank A is draining faster than Tank B because the slope of the graph for Tank A is steeper than that of the graph for Tank B.

The slope of the graph tells you the unit rate in gallons per minute.

Step-by-step explanation:

The slope of a graph is also known as the gradient of a line graph or surface. It is basically a number that shows the steepness and direction of a line graph or surface.

The steeper a line graph or a surface is, the faster an object falls from it.

From the graph, Tank A is steeper than Tank B, therefore, gas will drain faster from Tank A than Tank B.

Since gas is measured in gallons, the slope of the graph therefore tells you the unit rate in gallons per minute.

IrinaVladis [17]3 years ago

3 0

Answer:

Step-by-step explanation:

Faster than, slope,greater than

Slope, gallons per minute

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John, Sally, and Natalie would all like to save some money. John decides that it

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Answer:

Part 1) John’s situation is modeled by a linear equation (see the explanation)

Part 2) $y=100x+300$

Part 3) $\$12,300$

Part 4) $\$2,700$

Part 5) Is a exponential growth function

Part 6) $A=6,000(1.07)^{t}$

Part 7) $\$11,802.91$

Part 8) $\$6,869.40$

Part 9) Is a exponential growth function

Part 10) $A=5,000(e)^{0.10t}$ or $A=5,000(1.1052)^{t}$

Part 11) $\$13,591.41$

Part 12) $\$6,107.01$

Part 13) Natalie has the most money after 10 years

Part 14) Sally has the most money after 2 years

Step-by-step explanation:

Part 1) What type of equation models John’s situation?

Let

y ----> the total money saved in a jar

x ---> the time in months

The linear equation in slope intercept form

y=mx+b

The slope is equal to

$m=\$100\ per\ month$

The y-intercept or initial value is

$b=\$300$

$y=100x+300$

therefore

John’s situation is modeled by a linear equation

Part 2) Write the model equation for John’s situation

see part 1)

Part 3) How much money will John have after 10 years?

Remember that

1 year is equal to 12 months

$10\ years=10(12)=120 months$

For x=120 months

substitute in the linear equation

$y=100(120)+300=\$12,300$

Part 4) How much money will John have after 2 years?

Remember that

1 year is equal to 12 months

$2\ years=2(12)=24\ months$

For x=24 months

substitute in the linear equation

$y=100(24)+300=\$2,700$

Part 5) What type of exponential model is Sally’s situation?

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$P=\$6,000\\ r=7\%=0.07\\n=1$

substitute in the formula above

$A=6,000(1+\frac{0.07}{1})^{1*t}\\ A=6,000(1.07)^{t}$

therefore

Is a exponential growth function

Part 6) Write the model equation for Sally’s situation

see the Part 5)

Part 7) How much money will Sally have after 10 years?

For t=10 years

substitute the value of t in the exponential growth function

$A=6,000(1.07)^{10}=\$11,802.91$

Part 8) How much money will Sally have after 2 years?

For t=2 years

substitute the value of t in the exponential growth function

$A=6,000(1.07)^{2}=\$6,869.40$

Part 9) What type of exponential model is Natalie’s situation?

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$P=\$5,000\\r=10\%=0.10$

substitute in the formula above

$A=5,000(e)^{0.10t}$

Applying property of exponents

$A=5,000(1.1052)^{t}$

therefore

Is a exponential growth function

Part 10) Write the model equation for Natalie’s situation

$A=5,000(e)^{0.10t}$ or $A=5,000(1.1052)^{t}$

see Part 9)

Part 11) How much money will Natalie have after 10 years?

For t=10 years

substitute

$A=5,000(e)^{0.10*10}=\$13,591.41$

Part 12) How much money will Natalie have after 2 years?

For t=2 years

substitute

$A=5,000(e)^{0.10*2}=\$6,107.01$

Part 13) Who will have the most money after 10 years?

Compare the final investment after 10 years of John, Sally, and Natalie

Natalie has the most money after 10 years

Part 14) Who will have the most money after 2 years?

Compare the final investment after 2 years of John, Sally, and Natalie

Sally has the most money after 2 years

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4 years ago

The cost, C, in millions of dollars of producing T million tons of steel is modeled by the equation C = 220T + 1890.

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C = 220T + 1890.

Solve the equation for T.

220T = C - 1890

T = C/220 - 8.6

The steel produced is expected to be sold at a price of $310 per ton.

310 $/ton is a rate or slope. Write a linear equation where x is tons of steel produced and y is selling price of the steel.

y = 310x

Write and solve an equation to find the amount of steel produced if the selling price is equal to the cost of production.

* Here, note that the cost of production and tons of steel in the first equation is in the millions. The equation we just wrote for the selling price was in x tons of steel. This only matters in regards to the units you specify because; million/million = 1

The unit multiplier of all variables must be specified as same. Either everything is in millions or not.

Here, I'll leave everything in millions, change x (tons of steel) to T (mill tons steel) and "y" to "S" in million dollars selling price.

S = 310T

Set equal to Cost equation.

220T + 1980 = 310T

Solve for T, million tons of steel produced.

1980 = 310T - 220T

1980 = 90T

T = 1980/90

T = 22 million tons steel produced

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